Question

A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are oriented parallel to a uniform magnetic field of strength 0.037 T, the other two sides being perpendicular to the magnetic field. The magnitude of the torque on the loop is _____.

Answer 1

0.00185 Nm

Step-by-step explanation:

Parameters given:

Width of loop = 10cm = 0.1m

Length of loop = 20cm = 0.2m

Current, I = 2.5 A

Magnetic field strength, B = 0.037 T

Magnetic torque, τ, is given as:

τ = N * I * A * B * sinθ

Where N = Number of turns (in this case 1)

θ = angle between magnetic field and perpendicular of loop (90°)

A = area of loop = length * width = 0.2 * 0.1 = 0.02 m²

τ = 1 * 2.5 * 0.02 * 0.037 * sin90

τ = 0.00185 Nm